JEE Mathematics Limits Indeterminate Forms MCQs Set 05

Mathematics Objective Questions and Answers: Limits Indeterminate Forms

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Multiple Choice Questions

Question 1: \( \lim_{x\to1}\frac{(\sqrt x-1)(2x-3)}{2x^2+x-3} \) is equal to
(a) \( \frac{1}{10} \)
(b) \( -\frac{1}{10} \)
(c) 1
(d) none of these
Show Answer & Explanation

Answer: (b) \( -\frac{1}{10} \)

Explanation:
1. 2x² + x − 3 = (2x + 3)(x − 1) and √x − 1 = (x − 1)/(√x + 1).
2. The limit becomes (2x − 3)/[(√x + 1)(2x + 3)] = (−1)/(2·5) = −1/10.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 2: \( \lim_{x\to2^+}\frac{\sqrt{x-2}+\sqrt x-\sqrt2}{\sqrt{x^2-4}} \) is equal to
(a) \( \frac12 \)
(b) 1
(c) 2
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac12 \)

Explanation:
1. Split: √(x−2)/√((x−2)(x+2)) = 1/√(x+2) → 1/2.
2. (√x − √2)/√(x² − 4) = (x − 2)/[(√x + √2)√(x−2)√(x+2)] → 0.
3. The limit is 1/2.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 3: \( \lim_{x\to\infty}\left(\sqrt{x+\sqrt x}-\sqrt x\right) \) is equal to
(a) 1
(b) 0
(c) \( \frac12 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac12 \)

Explanation:
1. Rationalise: the expression is √x/(√(x + √x) + √x).
2. Divide by √x: 1/(√(1 + 1/√x) + 1) → 1/2.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 4: \( \lim_{n\to\infty}\frac{a^n+b^n}{a^n-b^n},\ a>b>1 \) is equal to
(a) −1
(b) 1
(c) 0
(d) none of these
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. Divide by aⁿ: (1 + (b/a)ⁿ)/(1 − (b/a)ⁿ).
2. Since 0 < b/a < 1, (b/a)ⁿ → 0, so the limit is 1.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 5: \( \lim_{x\to0}\frac{3^x-1}{\sqrt{x+1}-1} \) is equal to
(a) log_e 9
(b) log_e 3
(c) 0
(d) 1
Show Answer & Explanation

Answer: (a) log_e 9

Explanation:
1. Rationalise the denominator: (√(x+1) − 1) = x/(√(x+1) + 1).
2. The limit = [(3ˣ − 1)/x]·(√(x+1) + 1) → ln 3 · 2 = ln 9.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 6: \( \lim_{n\to\infty}\frac{4^{1/n}-1}{3^{1/n}-1} \) is equal to
(a) log₄ 3
(b) 1
(c) log₃ 4
(d) none of these
Show Answer & Explanation

Answer: (c) log₃ 4

Explanation:
1. Put h = 1/n → 0.
2. (4ʰ − 1)/(3ʰ − 1) → ln 4/ln 3 = log₃ 4.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 7: \( \lim_{x\to\infty}\frac{\cos x+\sin x}{x^2} \) is equal to
(a) 1
(b) 0
(c) ∞
(d) none of these
Show Answer & Explanation

Answer: (b) 0

Explanation:
1. The numerator is bounded by √2 in absolute value.
2. The denominator tends to infinity, so the limit is 0.

Teacher's Note:
1. Use the squeeze principle: bounded over unbounded tends to 0.

Question 8: \( \lim_{h\to0}\left\{\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\} \) is equal to
(a) \( \frac{1}{12} \)
(b) \( -\frac43 \)
(c) \( -\frac{16}{3} \)
(d) \( -\frac{1}{48} \)
Show Answer & Explanation

Answer: (d) \( -\frac{1}{48} \)

Explanation:
1. ∛(8 + h) = 2(1 + h/8)^(1/3) ≈ 2 + h/12.
2. 1/(h(2 + h/12)) = (1/(2h))(1 − h/24 + …).
3. Subtracting 1/(2h) leaves −1/48.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 9: \( \lim_{n\to\infty}\frac{n^p\sin^2(n!)}{n+1},\ 0(a) 0
(b) ∞
(c) 1
(d) none of these
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. 0 ≤ sin²(n!) ≤ 1, so the expression lies between 0 and nᵖ/(n + 1).
2. Since p < 1, nᵖ/(n + 1) → 0, hence the limit is 0.

Teacher's Note:
1. Squeeze between 0 and nᵖ/(n + 1).

Question 10: \( \lim_{x\to0}\frac{(1+x+x^2)-e^x}{x^2} \) is equal to
(a) 1
(b) 0
(c) \( \frac12 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac12 \)

Explanation:
1. eˣ = 1 + x + x²/2 + …
2. The numerator is x²/2 + higher terms, so the limit is 1 − 1/2 = 1/2.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 11: \( \lim_{x\to0}\left\{\frac{\log_e(1+x)}{x^2}+\frac{x-1}{x}\right\} \) is equal to
(a) \( \frac12 \)
(b) \( -\frac12 \)
(c) 1
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac12 \)

Explanation:
1. Combine: [ln(1 + x) + x² − x]/x².
2. ln(1 + x) = x − x²/2 + …, so the numerator ≈ x²/2.
3. The limit is 1/2.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 12: \( \lim_{n\to\infty}(0.2)^{\log_{\sqrt5}\left(\frac14+\frac18+\frac1{16}+\cdots\ \text{to }n\text{ terms}\right)} \) is equal to
(a) 2
(b) 4
(c) 8
(d) 0
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. The sum tends to 1/2.
2. (0.2)^(log_√5 (1/2)) = 5^(−2 log₅(1/2)) = (1/2)^(−2) = 4.

Teacher's Note:
1. Change log_√5 to 2 log₅, then use a^(log_a N) = N.

Question 13: \( \lim_{n\to\infty}S_n,\ \text{where } S_n=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\cdots\text{ to }n\text{ terms} \) is equal to
(a) \( \frac13 \)
(b) 3
(c) \( \frac14 \)
(d) ∞
Show Answer & Explanation

Answer: (a) \( \frac13 \)

Explanation:
1. The rth term is 1/((3r − 2)(3r + 1)) = (1/3)[1/(3r − 2) − 1/(3r + 1)].
2. The sum telescopes to (1/3)[1 − 1/(3n + 1)] = n/(3n + 1).
3. The limit is 1/3.

Teacher's Note:
1. Split each term into partial fractions.

Question 14: Let the rth term of a series be tᵣ = r/(1 + r² + r⁴). Then \( \lim_{n\to\infty}\sum_{r=1}^{n}t_r \) is
(a) \( \frac14 \)
(b) 1
(c) \( \frac12 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac12 \)

Explanation:
1. 1 + r² + r⁴ = (r² − r + 1)(r² + r + 1), so tᵣ = ½[1/(r² − r + 1) − 1/(r² + r + 1)].
2. The sum telescopes to ½[1 − 1/(n² + n + 1)].
3. The limit is 1/2.

Teacher's Note:
1. Factorise 1 + r² + r⁴ as a difference of squares: (r² + 1)² − r².

Question 15: Let a = min{x² + 2x + 3, x ∈ R} and \( b=\lim_{\theta\to0}\frac{1-\cos\theta}{\theta^2} \). The value of \( \sum_{r=0}^{n}a^r b^{n-r} \) is
(a) \( \frac{2^{n+1}-1}{3\cdot2^n} \)
(b) \( \frac{2^{n+1}+1}{3\cdot2^n} \)
(c) \( \frac{4^{n+1}-1}{3\cdot2^n} \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac{4^{n+1}-1}{3\cdot2^n} \)

Explanation:
1. x² + 2x + 3 = (x + 1)² + 2, so a = 2; b = 1/2.
2. The sum = Σ 2ʳ(1/2)ⁿ⁻ʳ = (1/2)ⁿ Σ 4ʳ = (1/2ⁿ)(4ⁿ⁺¹ − 1)/3.

Teacher's Note:
1. Find a by completing the square and b by the standard limit.

Question 16: \( \lim_{x\to0}\frac{x-\tan x}{x\tan x} \) is equal to
(a) 0
(b) 1
(c) \( \frac12 \)
(d) none of these
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. x − tan x ≈ −x³/3 and x tan x ≈ x².
2. The ratio ≈ −x/3 → 0.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 17: \( \lim_{x\to0^+}\sqrt{\frac{x-\sin x}{x+\sin^2x}} \) is equal to
(a) 1
(b) 0
(c) ∞
(d) none of these
Show Answer & Explanation

Answer: (b) 0

Explanation:
1. x − sin x ≈ x³/6 and x + sin²x ≈ x.
2. The ratio ≈ x²/6 → 0, so the root tends to 0.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 18: \( \lim_{x\to0}\frac{\log_e\cos x}{x^2} \) is equal to
(a) \( -\frac12 \)
(b) \( \frac12 \)
(c) 0
(d) none of these
Show Answer & Explanation

Answer: (a) \( -\frac12 \)

Explanation:
1. cos x ≈ 1 − x²/2, so ln cos x ≈ −x²/2.
2. The limit is −1/2.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 19: \( \lim_{x\to0}\frac{\sin^{-1}x-\tan^{-1}x}{x^2} \) is equal to
(a) \( \frac12 \)
(b) \( -\frac12 \)
(c) 0
(d) ∞
Show Answer & Explanation

Answer: (c) 0

Explanation:
1. sin⁻¹x = x + x³/6 + …, tan⁻¹x = x − x³/3 + …
2. The numerator ≈ x³/2, so the ratio ≈ x/2 → 0.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 20: \( \lim_{x\to0}\frac{1-\cos x}{x(2^x-1)} \) is equal to
(a) \( \frac12\log_2e \)
(b) \( \frac12\log_e2 \)
(c) 1
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac12\log_2e \)

Explanation:
1. 1 − cos x ≈ x²/2 and x(2ˣ − 1) ≈ x² ln 2.
2. The limit is 1/(2 ln 2) = ½ log₂ e.

Teacher's Note:
1. Identify the indeterminate form first.
2. Apply standard limits or expansions.

Question 21: If f(4) = 4 and f′(4) = 1, then \( \lim_{x\to4}\frac{2-\sqrt{f(x)}}{2-\sqrt x} \) is equal to
(a) 0
(b) 1
(c) −1
(d) none of these
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. Both numerator and denominator tend to 0; apply L'Hôpital's rule.
2. Ratio of derivatives: [−f′(x)/(2√f(x))]/[−1/(2√x)] = f′(x)√x/√f(x).
3. At x = 4: 1·2/2 = 1.

Teacher's Note:
1. The form is 0/0.
2. Differentiate numerator and denominator.

Question 22: The graph of y = f(x) has a unique tangent at the point (a, 0) through which the graph passes. Then \( \lim_{x\to a}\frac{\log_e\{1+6f(x)\}}{3f(x)} \) is
(a) 1
(b) 0
(c) 2
(d) none of these
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. f(a) = 0 and f is continuous (unique tangent), so f(x) → 0 as x → a.
2. ln(1 + 6f)/(6f) → 1, so the limit is 6/3 = 2.

Teacher's Note:
1. Use ln(1 + u)/u → 1 as u → 0.

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